Print all Longest dividing subsequence in an Array

Difficulty Level : Expert
Last Updated : 03 Jul, 2020

Given an array arr[] of N integers, the task is to print all the element of Longest Dividing Subsequence formed by the given array. If we have more than one sequence with maximum length then print all of them.

Examples:

Input: arr[] = { 2, 11, 16, 12, 36, 60, 71 }
Output:
2 12 36
2 12 60
Explanation:
There are two subsequence with maximum length 3.
1. 2, 12, 36
2. 2, 12, 60
Input: arr[] = { 2, 4, 16, 32, 64, 60, 12 };
Output:
2 4 16 32 64
Explanation:
There is only one subsequence with maximum length 5.
1. 2 4 16 32 64

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Declare a 2D array LDS[][] which stores the Longest Dividing Subsequence.
Traverse the given array arr[] and find the longest dividing subsequence till current element by using the approach discussed in this article.
Traverse the given LDS[][] array, and print all the elements of the sequence with maximum length.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
  
#include "bits/stdc++.h"
using namespace std;
  
// Function to print LDS[i] element
void printLDS(vector<int>& Max)
{
  
    // Traverse the Max[]
    for (auto& it : Max) {
        cout << it << ' ';
    }
}
  
// Function to construct and print Longest
// Dividing Subsequence
void LongestDividingSeq(int arr[], int N)
{
    // 2D vector for storing sequences
    vector<vector<int> > LDS(N);
  
    // Push the first element to LDS[][]
    LDS[0].push_back(arr[0]);
  
    // Interate over all element
    for (int i = 1; i < N; i++) {
  
        // Loop for every index till i
        for (int j = 0; j < i; j++) {
  
            // if current elements divides
            // arr[i] and length is greater
            // than the previous index, then
            // insert the current element
            // to the sequences LDS[i]
            if ((arr[i] % arr[j] == 0)
                && (LDS[i].size() < LDS[j].size() + 1))
                LDS[i] = LDS[j];
        }
  
        // L[i] ends with arr[i]
        LDS[i].push_back(arr[i]);
    }
  
    int maxLength = 0;
  
    // LDS stores the sequences till
    // each element of arr[]
    // Traverse the LDS[][] to find the
    // the maximum length
    for (int i = 0; i < N; i++) {
        int x = LDS[i].size();
        maxLength = max(maxLength, x);
    }
  
    // Print all LDS with maximum length
    for (int i = 0; i < N; i++) {
  
        // Find size
        int size = LDS[i].size();
  
        // If size = maxLength
        if (size == maxLength) {
  
            // Print LDS
            printLDS(LDS[i]);
            cout << '\n';
        }
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 11, 16, 12, 36, 60, 71 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    LongestDividingSeq(arr, N);
  
    return 0;
}

Python3

# Python3 program for the above approach
  
# Function to print LDS[i] element
def printLDS(Max):
      
    # Traverse the Max[]
    for it in Max:
        print(it, end = " ")
  
# Function to construct and print 
# Longest Dividing Subsequence
def LongestDividingSeq(arr, N):
      
    # 2D vector for storing sequences
    LDS = [[] for i in range(N)]
  
    # Push the first element to LDS[][]
    LDS[0].append(arr[0])
  
    # Interate over all element
    for i in range(1, N):
          
        # Loop for every index till i
        for j in range(i):
  
            # If current elements divides
            # arr[i] and length is greater
            # than the previous index, then
            # insert the current element
            # to the sequences LDS[i]
            if ((arr[i] % arr[j] == 0) and
                (len(LDS[i]) < len(LDS[j]) + 1)):
                LDS[i] = LDS[j].copy()
  
        # L[i] ends with arr[i]
        LDS[i].append(arr[i])
  
    maxLength = 0
  
    # LDS stores the sequences till
    # each element of arr[]
    # Traverse the LDS[][] to find 
    # the maximum length
    for i in range(N):
        x = len(LDS[i])
        maxLength = max(maxLength, x)
  
    # Print all LDS with maximum length
    for i in range(N):
  
        # Find size
        size = len(LDS[i])
  
        # If size = maxLength
        if (size == maxLength):
  
            # Print LDS
            printLDS(LDS[i])
            print()
  
# Driver Code
arr = [ 2, 11, 16, 12, 36, 60, 71 ]
N = len(arr)
  
# Function call
LongestDividingSeq(arr, N);
  
# This code is contributed by ANKITKUMAR34

Output:

2 12 36 
2 12 60

Time Complexity: O(N²)
Auxiliary Space: O(N²)

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